Core saturation and flux density calculations

[machinehead]

Hello everyone

I had built a small SMPS years ago, that puts out 15V 0.3A from the 120VAC line, and I was now trying to upgrade it to give twice the power (0.6A).

The core ( a 12mm ferrite toroid ) starts to go into saturation at powers just above the original rated power. I have read in some places that increasing the number of turns can raise the saturation threshold (perhaps by lowering the core flux density?).

So I'm considering increasing the primary turns from 100 to 200 (and the secondary accordingly) to hopefully keep this trafo from saturating at the higher poer levels. Is this assumption correct?

I'm a bit confused by the theory, which states that:
B, flux density = [Volts] [Seconds] / [Meters squared] [Turns]
so, more turns, less flux density, OK so far.
but
H, magnetomotive force = [Amps] [Turns] / [Meters]
so, more turns increases H

But B and H are inexorably tied by the B-H curve, so increasing H will bring B higher along on the curve, resulting in higher B, not lower B. The two equations seem conradictory.

I'm sure math doesn't contradict itself, so what part of this am I getting wrong?

thanks

Pamela

[jackinnj]

make your life simple -- go to your town dump, get a beat up ATX supply -- take the ER core out -- roast it in the oven (wrapped in aluminum foil -- 400 degrees for 45 minutes) and then gently pry apart -- taking care not to break the bobbin or damage the two pieces --

this core should be capable of several hundred watts and is quite compact.

if you have LABView (or can get a student edition) you can make a BH Curve tracer with a couple resistors and one capacitor -- it's in the most recent issue of EDN (www.ednmag.com). that way you can characterize the core without scrambling to find a manufacturer's data sheet (which you probably don't have anyway.)

i have posted a very simple and very acurate impedance measuring unit with which you can measure the inductance of the primary winding -- search under STEBER -- the fellow at Illinois University who wrote the software --

[poobah]

The first equation has seconds in the numerator. This tells you that increasing the turns will increase the inductance and require longer (more seconds) to reach a given flux density.

So there is no contradiction.

First law of engineering... nothing is free. You may need a bigger core.

Core size is proportional to [L*I^2] / 2 This is the kinetic energy the core can "hold".

You didn't mention your topology or what inductor/transformer you are talking about. You may be able to increase your switching frequency to increase power.

Check application note #19 from Linear Technologies... specific to their part of course... but packed with info and equations.

[jackinnj]

oh, forgot -- for push-pull applications (continuous) you can use this spreadsheet on my webiste:
http://www.tech-diy.com/smps_xfmr.xls

[machinehead]

Thanks

I will follow up on that Linear tech note.

I don't have a shortage of larger cores. But this thing needs to fit into a constrained pre-existing space. My original unit is about the size of a pack of gum.

[machinehead]

Oh, yes..

the primary inductance is 1.1mH. Topology is forward with RCD reset. The 12mm toroid *is* the power transformer. The whole circuit is 2" x .75" x.75", smaller than a pack of gum, actually.

I know about using a larger core, but I thought squeezing 10W instead of 5W from that core might not be such a big deal to ask of it.

I mean, is it true that adding more turns can sometimes solve a saturation problem, or is that false?

thanks

Pamela

[poobah]

Increase your frequency and remove turns.

Look at the math...

[Eva]

You all are overlooking a simple fact:

In forward and push-pull converters, power output is not related to transformer saturation at all. If the transformer starts saturating above a certain power level, then there is something really wrong with the circuit. The only limiting factor in these circuits is output inductor saturation and I^2*R losses.

I'm a bit confused by the theory, which states that:
B, flux density = [Volts] [Seconds] / [Meters squared] [Turns]
so, more turns, less flux density, OK so far.
but
H, magnetomotive force = [Amps] [Turns] / [Meters]
so, more turns increases H

Given a fixed frequency and pulse length:

Increasing turns decreases B proportionally to N'/N and increases magnetising L proportionally to N'^2/N^2. This means that the magnetising current gets also reduced by a N^2/N'^2 factor.

So if we double N, in the H forumla '[Amps]' gets reduced to 1/4 and '[Turns]' gets doubled, so H is halved.

[machinehead]

Eva:

Thank you for addressing my misunderstanding head-on.

So adding turns lowers B, while increasing H to a *LESSER* extent, still resulting in a net drop in flux density.

I guess the price you pay is higher leakage inductance from the higher N...

Thanks again
Pamela